493 83 86MB
English Pages [130] Year 2021
…
●
●
●
●
●
●
●
●
●
● ●
●
15
16
17
18
More Examples
I Methanol
H
a
H
3 NH Ammonium ion
removed
one e is
from 5 valence e
N to make it
g
positive charged
of
Lewis structures 2
CHIH
i
f
19
Ufo
Methanol or formaldehyde unpaired
H O
n
n
H
H me
T
Makingone bond
Cand O
between
leaves one unpai
e on each c
ii
fi
O
of CandO they pairupto form a second bond o
4
5
20 6
Ctej Methylanion inning
a
fg
is
O e
put onto the C
UfNOz
ii
Nitromethane
Unstable
n.fi X H I
H c
Uycoo n
l H
I
Acetate ethanoate ion
µ
O H
N
i
o
N has toe
Q
ms
octetexceeded
i
Not this
Polar Covalent Bonds & Dipoles 1 21
When a covalent bond is formed between atoms of different electronegativity, the bonding pair of electrons is not
shared equally but is attracted towards the more electronegative atom. Such a bond is said to be polarized , or
described as a polar bond, and the charge separation is represented by symbols δ− and δ+ which represent partial
negative and positive charges. Alternatively, we show the direction of bond polarity by an arrow with the
arrowhead pointing toward the negative end and a plus sign on the tail of the arrow at the positive end.
The bond is ionic only when the electronegativity difference is appreciable (typically, >1.8 on the Pauling scale);
otherwise, the bond is covalent.
Bond Dipole Moments 22
Bondpolarities can rangefrom nonpolarcovalent through polarcovalent to totally ionic eg
0
Cte Cte MHz OH Cl Uhs N 6 Ctf Utz Utz
Ionic
NonPolar
Increasing Polarity
an individual bond is measured as its bond dipole moment µ Polarity
St S amount charge at eitherendofdipole µ 8xd
H H U D distance bw the charges
SI unit Debye D k d
30 Coulombmeters
I Debye 3.34 10
a protonand e 9 1.60 1019c were 10 m apart the dipole moment wouldbe
74
1.60 1029 cm 1.60 10 4 8D D
µ 8 0 4.60 1019c x loWm 3 34 1030
H N C N H O C O
c f C y µ c
0.2213 0.86D 1051 D 1.56D 0.3 D 1.3113 l 53 D
of
Seg
of
If
292
8xd
23
Molecular Dipole Moments rholeculardipolemoment is the dipole moment of the molecule taken as a whole molecule's overall polarity Its a good indicator Moleculardipolemoment
10N a
eg
120
is the vector sum of individual bond dipolemoments
1All
forceon
this sphere
1240,11200
loaf
of
10N
zero became 3 forces Vectorsum O
up H
C
O
H
c
O
µ 0
Non Polar
µ 2 3D
Molecule
polar molecule
IF
0
similarly
F
g hi
L
14 0
p
If
µ o Nffjepolate
Hisc pH c
F
H
B
H
Non Polar
Molecule
Bfd
Br µ lo9D
Polar molecule
U y
c
4
U U
U
Threepolarbonds polarbonds fourPolarbonds Two
cancel dipoles All All dipolescancel
bothdipolescancel _O MR
Mr 0 µr O
Label thepolarbonds in eachmolecule Indicatethe direction
b Uesufocte
Clea d as a
Br
It
IF
If
401
24
It
at't
17dB
of net dipolecifany g
Bt
Br
MAB
e
a
Br
a
H C
1 s
s pigsty 9 1
U
d
U
s
qq.BR Br
Br
Mr_0
R
b
o
f II Ms cee
25
Br
7 Mr
c
It I
µ
l
µr O
H
Br
Bothdipoles are equal opposite
Benzene is a planar molecule and both Br are in the
plane of the benzene ring Both dipoles cancel each other
Br
and cancel
each other completely Br
e
11 l
Br
fl I
µ
H
I Mr
Bothdipolespoint in
the
same direction
hence add up togive a resultant dipole
26
Electronegativity allows us to predict the charge distribution in an organic molecule
s S 0 St S St S
I Cee CEN s Ctl OH
eg CSt 11111 C
µ H As µLd
The major factor Influencing the E N is the energy the orbitals that
to rightacross the periodictable the valence the atom uses to acceptelectrons As onemoves
orbitals become lower in energy withinthe same period Going down a group atoms get bigger
in size and valence orbitals are higher in energy E N increases as orbital energy decreases
Then He should have the highest E N as it has the lowestenergyvalenceorbitals but He
can't accept any more e b'coz its a noblegas The atomwith the lowest energyvalence orbitals
that is not a noble gas is F which is the most E N atom
I
I ayy
left
of
Electrostatic Potential Surfaces
Just for your knowledge no need to memorize
27
Cebu showing the polarity with St and 8 is veryeasy However in more complex molecules there will be manydifferenttypesof bondswithdiffering degrees of polarity and the overall molecule will reflect the sum ofthese Thesimple St S
In
a simple molecule
like HCl
or
symbolism is no longeradequate and we need an alternate way to view the charge distribution in complexorganic molecules
Example
Example
Example
Classify each bond as nonpolar covalent or polar covalent or state that ions are formed. (a) S 𑁋 H (b) P 𑁋 (c) C 𑁋 (d) C 𑁋 n F mH N Cl
M
Would a bond between each of the following pairs of atoms be covalent or ionic? (a) O, H (b) C, F (c) Li, F (d) C, Mg
Show the polarity of each of the following covalent bonds by partial charges and by a dipole arrow. (a) O–H (b) C–O (c) C–Mg (d) B–H
28
Example
Example
Example
Classify each bond as nonpolar covalent or polar covalent or state that ions are formed. 29 St St S s (a) S8Wa 𑁋 HSt (b) P 𑁋 H (c) C 𑁋 F (d) C 𑁋 Cl b m N EN of P and H Polar Polar Polar NonPolar are very close so covalent covalent covalent covalent there is hardlyany Polarityin P H bond Would a bond between each of the following pairs of atoms be covalent or ionic? (a) O, H (b) C, F (c) Li, F (d) C, Mg
d
DEN to i polarcovalent
de
DEN 1.7
butstill
DEI s 1.7
de
DEN 1.7 Ionic
i Polarcovalent
Polarcovalent as carbondoesn'tformionicbonds
Show the polarity of each of the following covalent bonds by partial charges and by a dipole arrow. (a) O–H (b) C–O (c) C–Mg (d) B–H
8
O
H
St
St
s
C O
s
s1
C Mg
St
S
B H
30
Group Electronegativities
Its often convenient to consider groups that make up particularportions of a molecule as having their own electronegativity compatible with Pauling Seale Group
Electronegativity
Group
Electronegativity
Utz
2.3
CECH
3.3
Ctfu
2.8
CEN
3.3
Cteclz
3.0
MHz
3.4
cuz
3.0
NHI
3.8
Cfg Ph
3.4
Moz
3.4
OH
307
Ctectez
3.0
3.0
Bonding Patterns & Formal charge 31
Before we start with calculation of formal charge, let’s just understand its importance in chemistry. Suppose you do
a lot of monetary transactions everyday, then how do you keep track of your account balance? You make a ledger.
Formal charge is also like keeping a track of valence electrons of an atom.
We will be studying lots of reactions and their mechanisms in this course and reactions involve movement of
electrons. When electrons are being moved during reactions, we need to keep a track of this in order to understand
the reaction mechanism.
Bonding Patterns & Formal charge
1 32
Assign formal charge
a
e
te
te Ye
ch E
33
to C N and O atoms in the following molecules b
f
µj teYg
es
y
g
ay
Uy
EN
d
H
pi
NEN
H
BL
a
e
solution
a
S1
34 H
I
He c i
H
teach
O
1H c
b
nµ
p
has 3 e oo
T
H
c
ye
T
has teoc.io
f
T
has 5e
bing.nqe.n.ie
N has 6 e i I formalcharge
o
0
p
1
H og H
d
Cte c N
has
e
o
haste
G
11
n
HY has5 e
hlasse
Ue
g
O
cteg.NO N
T
hasUe
d
Se
X has 55
d
I
L de
Se
35
Formal charge and Bonding Patterns
In organic chemistry we will be dealing with lots simple and complex molecules so its important of to have a proper understanding bonding patterns of common atoms like C N and O
of
Dueto 4 valence e C
EC
forms four bonds in Neutral state
c
c
A
octet is maintained
E
formal charge
formal charge 0
if
Y
and not
t.g.E
e.EE
I
E
These arrangements for formalcharge 1 are wrong as octetofcarbon is exceeding due to the
unpaired e
Patterns
Formal charge and Bonding
Due to 5 valence E
N forms 3 bonds
and keeps one lonepair
e fc o
Can't form 5 bonds as
octet will be exceeded N
and N doesn't have
empty d orbitas for that Re I
Similarly Oxygen valence 0
1 o 1 o o
0 O O
fC D fc I
hi
ni
36
N
htt
N
of
iii I
Fc
These bonding arrangements ensure octet completion
for
g
I
0 Fc
I
37 shapes organic Molecules
Molecular structure
Molecules have 3Dshapes What do theylook like Why does a moleculehave a particularshape
The search these answers led to the development various bondingtheories These theories
explained the known structures and also allowed chemists to predict structures newly
discovered molecules When a predicted structure was shown tobe wrong theory was refined until
experimental result it could account
Lewis concept concept localized e pair bond Coudritexplain the geometry bondangles
stability
V BT Explains formation bonds through overlap atomic orbitals
vs EPR theory Explains the right arrangement bondpairs and tone pairs in space around
central atom to achieve stability
correct
Allows us to modify the atomic orbitals in such a Hybridization way to explain
bond angles in molecules Assigns geometry bondangles and bond lengths
of
for
of
of
for
of
of
of
of
Knowledge properties
ofmolecularstructures is critical to understanding of their
Virtually every
molecular structure
physical chemical great extent on the shapes of acidity basicity reactivity ele depend on
biochemical process hinges to a
interacting molecules Properties
like smell flavour
VSEPR Valence shell e pair repulsion
e pairs mean
bondpairs
lonepairs
g
Based on simplefact that e pair repeleachother and want to stay as far awayfrom each that other as possible Electronpairs around central atom arrangethemselves in such a repulsions are
I
38
VS EPR theory
minimum and structure is stable
way
Tetrahedral shapes
z
Pointed
away pointed towards You
fromyou
Y'npitamnesame
a
It
39
µ.gl
or
does amully
5
Pointed
Ball stickModel
you
away fromyou Tetrahedral is one the most of
3D formula
txt
Pointed towards
9
Lewisstructure
or
common geometries
you will
study in organic chemistry Visualize it perfectly
spacefillingmodel
Tetrahedral and Trigonal planar geometries
e
e
z
C
o
1
Ammonia
NH
Pyramidal
shape
Geometry Tetrahedral
Nater Dihydrogenmonoxide
HD
Shape Pyramidal
Tetrahedral
Geometry
Ehr
40
Linear Geometry y
41 S
linear
i
Hy
Ethene
Trigonal Planar Geometries
HCHO
Methanol
Koryo
U
5092
ThionylChloride
Tetrahedral
Seo
42
i
P
6
1
p 304
U
POU
p
a
U
U
Nyo HMOs
Pce
phosphorylchloride
Phosphorous
Tetrahedral
Tetrahedral
trichloride
HO
yo
NitricAcid TrigonalPlanar
Doo
HO
Nyo
HN0z NitrousAcid TrigonalPlanar
Atomic Orbitals and their shapes
43
An orbital of specified energy can bethought ofeither as the space around the
atom available to an e an e
of an
or as a mathematical formulafrom whichthe probability offinding
at a particular location may be calculated
nucleus
colours whythesedifferent
These two colours represent two different phases The phasing is solely a result the mathematical of colour indicates that the function is positive in this functions describing the orbitals One region of space and the other colour indicates it is negative
44 There is no other meaning to be given to these phases forinstance the probability of finding an e in the differently phased regions is the same
In atomictheory and quantummechanics an atomic orbital is a mathematicalfunction which describes the location and wave like behaviour of an e in an atom
What is the difference b w orbital
Have function contains all the
and
etc
4
4µm locate an e
wavefunction 4 dynamical info about the e ie its energy coordinates is characterised by 3 quantum numbers n eand m
Havefunction we
define its
a
has all the info about
e
but when we want to
n Land M and then it's called an orbital eg is
2s atomic orbital
42,0
Valence Bond theory
VBT
45
What is a covalent bond, and what characteristic gives it strength? And how can we explain molecular shapes based on the interactions of atomic orbitals? One very useful approach for answering these questions is based on quantum mechanics and called valence bond (VB) theory.
A covalent bond is formed whenorbitalsoftwo atoms
overlap and a
pairof e overlapof
occupy the overlap region In the terminology of quantum mechanics the two orbitals means their functions are in phase constructiveinterference
VB T assumes that all bonds are localized bonds ie bonding e are fixed b w two This is actually an invalid assumption combining atoms eg A B because w 2atoms many atoms bond using delocalized e e which are not fixed b
Two other key concepts of VBT were developed later For the formationof a bond btw 2 atoms aeet Attractiveforce
Met repulsivetree
e
t
2 Resonance 2 Orbital Hybridization e
Repulsiveforces Aetraceiveforces
Overlapping of Atomic orbitals
When two atoms come close to each other, there is overlapping of atomic orbitals. This overlap may be positive, negative or zero depending upon the properties of overlapping of atomic orbitals. The various arrangements of s and p orbitals resulting in positive, negative and zero overlap are depicted here.
Overlapb w same phases
1 PositiveOverlap ie constructive interference
Overlapbtw opp phases
2 Negative overlap ie destructive interference
No effective overlap
3 Zero overlap
orbitals
of
46
Natureof Covalent bonds SigmaBond
formed by axial
T or
and IT bonding
47
end toend overlap ofatomicorbitals
Leads to greater overlap b w orbitals and hence is a stronger bond
Is
Is
Internuclear axis
Two Hatoms
f
Overlapping of orbitalstakes placealong the internuclear axis
the axis connecting the nuclei combining atoms of
Overlapping
ie Greater e density in b w the nuclei
Of S orbitals o
or
H H
Natureof Covalent bonds IT Bond
formed
r
and 1T bonding
by parallel orsideways overlapofatomicorbits
Extentof overlap is less
Exampleof pit pit bonding
If Example of
us
so IT bond is weaker than
bond
s orbitals don't form it bonds
day
881 dry
2 lobes
only d orbital are in overlap
p of each participating
gfgffemanfiutoietibeasueo.es of each d orbital are PIT dit bonding
participating
in overlap
formation of Ctly molecule
4g
as per VBT
A bond is formed between two atoms when both contribute one electroneach ie At 1B AMB Two half filled orbitals overlap to
1B Dl
G's configuration of C
Es
C
IS
Dt
25
ftp
off
give a bonded
2P
t H
can form only 2 bonds in ground state
9 9 9 t
pair
It hasonly 2 halffilledorbitals so it
4 f f
e
t
t
H H H
One e is excited from as Ip to give 4 unpairedorbitals toform 4bonds
4 halffilledorbitalsof Cand 4H overlap
Onebond is formed by Is 25 overlap threebonds are Is ap overlap
This means that all 4 bonds are not same 3 are similar and one is different
50
But it's
experimentally confirmed that all C H bonds of Ulu are alike So in situations like this normal VBT is not successfuland we introduce a new concept
called Hybridization
t.TL
BEBEE P orbital
If
Mixing
iir
EEEEo EESE EEEEEEEEOEESE.no orbital 3 sp orbitals p orbital 2
s
I mix 2 bowls rice with 1 bowl dal and dothe cooking
I get
palace
So different orbitals
rice and dal when mixedtogether give a new set of equivalentorbitals Cpulao
Hybridization is the process where orbitals
ofslightly differentenergies like s Pd
intermix
to generate a new set of orbitals which are equivalent in energies shapeand size known as
hybrid orbitals like Pulao in the aboveexample
52
Now let's discuss the formation of Ctly once again Is
c CGS C CES
25
ft ft off f
2p 9 t g g g
Ones andthree p orbitals mix to generate a new set of 4 hybrid orbitals sp'sorbitals
Hybridize
off
1
Is
These 4
Nowthese are
9 f f f t
Is
t
Is
no longer S or
p orbitals but are
sp3 orbitals having similarenergy and shape
t
Is
four hydrogen atoms ready to overlap with 4 sp3 orbitals of C atom to give Uty
sp orbitals axially overlap with Is orbitalsofH to form a
tetrahedral arrangement
52
No of hybrid orbitals formed Hybridization
Stp
No
of participating orbitals
Noof hybridorbitalsformed
Bond angle
Geometry
sp
2
Linear
St Ptp
3
TrigonalPlanar
1200
St
sp2 p Ptp sp
4
Tetrahedral
109.50
of S character
sp3
sp
sp
501
33.39
Electronegativity
sp sp2 sp
Relative Overlap
sp
sp
sp
25
As
180
S orbital is closer to thenucleus the
greather the soloof a hybrid orbital
the lower its energy the smaller its size
p
higher its
Electronegativity
53 Hybrid orbitals undergo better overlap compared to pureatomicorbitals this results in stronger and more stable bonds Geometry of a molecule is governed by hybridization bonds and not IT bonds Hybrid orbitals form
O S
A
t
CA
t
CA
t
S
O s
of
t
t
Be
Be Be
Sp Hybrid Orbital
CS
D
Ry
CA Ry
t
A
B
As p character increases the hybridorbital becomeslonger
sp2 Hybrid Orbital
I
sp3 Hybrid Orbital
54 Totalenergy ofthe orbitals is not changed by the hybridization and energy each hybrid orbital of in a set is the mean of the energies of constituent Aos L
f
Unhybridized Orbitals
formation 55 Ethane
H H
I l
H C H c
I 1
H H
Hoo
bonds
Each C atom has
like Chey it needs hybrid orbitals
axis
This bond is symmetrical along the internuclear
C C Now I hold one carbon atom fixed and rotate the
the carbonatoms
other it will not affect the overlap b w
bond ie rotation about c c is allowed
in yourhands formed BY
can visualize this byholding a pen withcaps
sp3 sp overlap and rotate the pen while keeping the cap pen fixed
of
four Usps
I
T
T
X
just
µ
if
you
of
formation Ethane
c
eg
holdthepenbody
H H
Cii
µ H
y
H
H
what you
see
of
GHG
56
c
This pen can be considered an example of bond here
g
12 HC
th
11
b
f
TH
rotate
the cap
to Youdidn't faceany problem ie
Rotation about a single bond is allowed 2 we will use this concept in stereochemistry
OD
eye Carbon
pear n or
I
Goo
Hold the front
at
cartebaornfansdonrobtyategoo
front carbon
j
formation
of
Ethene
57
Hy
Each Carbonforms 4 bonds 2 bonds C H Require 3 orbitals is a bond C C Hybrid f it bond c C J Requires 1 p orbital
J
I se
be will excited
to2P
aroundState
I
these 3 orbitals will form
3 o bonds cloud is above the plane bond e IT halfof it molecule rest half is below
7 of
entire GHy molecule is
in one plane becamethe
shape of sp2 hybridization is trigonal planar Noofatoms in sameplane 6
formation
of
Ethene
58
Hy
2porbitals
t
b
H
H e
Overlap oftwosp2hybridorbitals
forms the C C
Overlap
r bond
of sp2and
torms the
c H
Is orbital bond
H
H
f
cu
T H
ITbond
c
c C
H
H
of two 2p orbitals formsthe bonds are in the plane the of
Overlap
C C itbond
5 molecule but ITbond extends above and below the plane ofthe molecule
e density of ITbond is not concentrated on the
axis joining the two nuclei
Because the e density in a 7Tbond is fartherfrom
the two nuclei IT bonds are usually weaker
59
Restricted Rotationabout IT bond
we have already learnt that rotation about bond is allowed Now let's understand what happens when we try to rotate about a T bond p orbitals ie sideways overlap IT bond is formed
of
by
As we start rotating
Rotate
f
gg
d
C
got
fixed the overlap blue p orbitals is
f
destroyed So rotationabout a it bond is restricted
d Cz
Fcf
by keeping a
Doublebonds are rigid and can't be
twisted
Rotateabout bond
bykeeping
one carbonfixed
of
Motagua
Neill use this concept
in geometrical
isomerism Sovisualise it clearly
60
Does it mean that we can never rotate about it bond
IT bond results from sidewaysoverlapof porbitals When the two porbitals are parallel the overlap is maximum If one of the carbons is rotated about the c c bond the 900 when there can be no overlap between P orbitals decreases and becomes zero at
1T bond In other words rotation about the does not take place at normal temperature
What is the meaning of
and
e
is a
high energy process which
in the lobes of orbitals
These lobes don't refer to positive or negative charges since bothlobesof an e cloud orbital must be negatively charged These are the signs of wave function 4 when two parts of an orbital are separated by a node 4always has oppsigns on the two sides of the node This also explains why s orbital has only or sign and not both like p orbitals
Ethyne Cs
formation
H C C H
180
carbon forms 4 bonds Each
I C H bond Requires 2
bond hybridorbitals I C c
2 C C ITbonds Requires 2
P orbitals
Hz
of
Gl
y
f
se be
will
excited
4 atoms are linear Two IT bonds extendout from the axisof
the linear molecule
1
62
Trick for calculating Hybridization 1800
A
TRICK for Hybridization
Calculate no
E 1200 v
109.50
3
I
bonds
free lonepairs
Moreclarity on free lonepairs will comeafter nanee so don't worry if youdon't YYntifgyur.es
2
of
T
1 Free L P
2
3
4
sp
linear
sp2 Trig Planar sp3
5
sp3d
6
sp d2
Tetrahedral
TBP Octahedral
This is an easiermethod Detailedtheory is covered in chemical Bonding topic and dsp2
hybridization is covered in coordination compoun
topic
Trick for calculating Hybridization
63
TRICK for Hybridization
Calculate no
bonds
free lone pairs
Moreclarity on free lonepairs will comeafter don't Yeventingthrees.onance so don't worry if you T
Free L P
2
3
of
4
sp
linear
sp2 Trig Planar sp3
5
sp3d
G
sp3d2
Tetrahedral
TBP Octahedral
Calculating Hybridization in
H H
eg µ c q
µ c c H
µ I l
c 4
i s sp3
sp2
Cee _UL
Cte
Ut Sp a
p
U
G
ay ay
ay
As
cha ausps
64
organic molecules c
H
H
to LD
f
I
c H
N H
3
n
sp2
H
2
sp O p
Utz dps
I
Io zLp y i
I
11
H
spz
11
20 10 L P
4
I LP
3
attacks 2
Ips
2LP 3
sp2
o
P y
sp2 40 06 5
U a
Fps
sps
ab
Ips
ut cha 2
at
Vsp
I
at
at
Tey
or
t
thatyaall care sp2
65
0 i
H or
Cte
30 0 up
stp3
Cee
or
y
CHP
i
µ
µqq
sp3
te
or
sp
ite
30 0LP
i sp2
the odd e isin pure p orbital
iii Vf
cee
Asp3
oteff
Vhf
Sps
Restallaresp
carbons
sp
dP
y
Restallaresps can
Sps
3ot2LP 4
E
Utz Slp Isps
sp2
o
µ
UF
O
o
08 spa0 o H
S OH 0 1 OH
To spa sp
µ O
sped
N
O 0
sp2
as CH CECH
Determine the hybridization around the highlighted atoms in each molecule
b
c
Ctf
c
Utz
Seldane is a major drug forseasonalallergies Relenza is a common antiviral
I a
p
b
66
t
I
b Seldane
O
OH OH
throat Tf OH NH NH
0
Draw all the lonepairs of e in bothstructures
fd
7 MHz
NH
Relenza
Which orbitals overlap to form the covalent bonds indicated by arrows a b c and d
67 as CH CECH
sp3 sp
b
NO
ttbcjcHz
pz ctezjprVsp yo
no
sp sp Seldane
I i
Ipa
T.N.ie µµ O
Tsp
Il
ao
I
c
fd
T.N.tl
1.7142
Ipa
spa sp sp2 sp2 Relenza
Place
possesses
it in the following structures
CEC
c
N
a
es
formal charge over any atom that
yO
b
H CEO
Q
Y
off NH
EJ't
US
Molecule
and
A
11
rub
N
Ctb
A contains
sp
carbons
d
UtzNH
t 0
I
Q
68
un
0
Ttb
Tan
t
4 Ctl
theobromine
sp2
Howmany lonepairs in theobromine
MHz
I
wane Il t µµTNXNHz 2 Melamine
Howmany lonepairs in
melamine
Solution to
Practice Problems
a
es
CEC
0
b
69
t
y o.info Ill
yo
I
s
u
O
IX
NH
Hes
Molecule
and
A
N
A contains
6 sp's
carbons
0
8 sp2
Q
d
CtlzNHz
EI O
ab
ab
N
c
H CEO
Hq
Il X
II Mil
its No t
Uts Lonepairs _8
ivies in
I X
qq.HN
N
iiiiq
Lone pairs _6
Quick Recap of MOT
70
MOT is very important theory to understand the phenomenon which involve molecular energy covered levels Its also based on quantum mechanical model atom and you must have of this in detail during chemical bonding getback here
If not
then brushup the basicsofMOTand then
Atomicorbitals Aos combine to give Mos
In phaseinteractions out ofphase
constructive interference
destructive interference
Bonding MO Antibonding
orbitals that contain pairof e which are not bonding bond Axial overlap sideways overlap
IT bond
Brno
MO
ABMO
or antibonding
higherenergy lower energy
Nonbonding orbital
BondingOrbital
Antibonding Orbital
of
fo
Sigma overlap
Dt
f f t
f
it overlap
Molecular Orbitals 72
According to molecular orbital (MO) theory, when two AOs overlap, they combine to form two new orbitals called
molecular orbitals (MOs), and the electrons from the original AOs are now accommodated in the new MO(s).
Whereas AOs are localized on atoms, MOs are associated with molecules and sometimes extend over a whole
molecule.
The simplest example is the formation of the H–H bond
of a hydrogen molecule (H ), which involves the 2
interaction of the singly occupied 1s AOs of two H atoms.
When the two AOs of equal energy overlap, two MOs are
produced. One is lower in energy than the original AOs,
while the other is higher and the new MO of lower
energy becomes occupied by the two electrons (as per
Aufbau Rule) which were originally in the two AOs.
It is this pair of electrons which constitutes the covalent
bond . The H2 molecule is lower in energy (more stable)
than the two separate H atoms, and the energy
difference, which is liberated as heat, is the bond energy
of the H2 molecule.
Molecular Orbitals
why Bonding MO is more stable Because
is greater which reduces the repulsion b w
e cloud
btw 2 nuclei
73
in BMO e density blew the nuclei nuclei and provides stability
Molecular Orbitals
why Antibonding MO are unstable
and hence repulsion b w the nuclei
74
In ABMO the
e density b w
the two nuclei is zero
is maximum This makes the ABMO Unstable
Molecular Orbitals
c
In a 1T bond the 2 e
and ABM0
stay in BMO
is empty
this empty DBrno will be
used in upcoming chapters
concepts like
to define etc
hyperconjugation in alkene
a
y
O
o
0
75
Bond strength
Bond length and
atoms comecloser to form a covalent
As 2 bond potentialenergy the system
decreases and reaches a minimum Point3
This point represents max stability
lowest energy and distance b w the
atoms at this point is called Bond
length
atoms comecloser than point3
the repulsiveforces b w their e clouds
become stronger than attractiveforces
system becomes unstable This
pushes the two atoms apart towards
3
point again
Bondlength
H H
of
If
of
76
77
4 Bond length 9 as we move down the group as distance b w
the nuclei 4
Bond
Energy The strength a covalent bond depends on the
the attraction b w the nuclei and shared e
magnitude of
The energy required to break the bond a particulartype is
known as bond energyCB E
Bored its an endothermic process breaking always requiresenergy
heat is
9 A BCgs absorbed Cgs A AH se KTImol
Bond energy
Greater the B E
strongerthe bond
of
of
Big
Average BondEnergies Kamel and bond lengths pm
Noneed to memorise gust for your reference
78
Bond length and Bond strength
2g BondOrder No bonds b w 2 atoms of
Increasing Bond length
g
yeah Increasing Bond Strength
Bonding e
c
eg
N
c
2
c
c
BO I BO 3 B0 _2
As bondorder 9 increases bond length
and bond becomes stronger
All C C don't have some bond length
as hybridization plays an importantrolehe
Clsp3 c ClspDtclsp2
Utz Utz eg Us at _Utz
1 53 1.55 Ao 1 Ug 1.52 Ao
same trend we find in c H bondlengths
sp
Antibonding e
in the table
As
Bond length and Bond strength H
UtsUta H
f
Sp
80
I 111
PM
Ctf_qµ
This again confirms that s
f
spz
110pm
Asthe soloof carbon 4
Cte C H
I
sp
electrons are closer to
109PM
the nucleus
in the hybridization the bond length d
81
82 shorter
sp
f
a
sp 3
shorter Shorter
I
Shorter
sp
Br as
Br
II
pc
IIH
µ
PolarMolecule
b
IF
th
CpBr
H
Polar Molecule
F
IT
C
F
gt F
MR 0
sp3 4
p d's
f
c
U
Y
PolarMolecule
te e
UH
c
U H
µr
O
83
Polarizability a
molecules that
very important property of
youmust havestudied during
Rajan's rules in
mical bonding let's have a quick recap ofthis as well
they respond to electricfields In particular e in molecules are so their positions can shift to different extents when any external
Since e are charged particles
mobile to varying degrees
electric field dueto a charge is nearby
g
e e
e
e
e
a simple representation
of a neutral atom
e
Red
Blue
recharge e
cloud
e density has shifted duetothe presence of ve charge nearby This is called polarization ie development of ve
It has a symmetrical
distribution of charge in the absence
ofany external electricfield
St
e
Place a
cation
near this
atom
s e
e
e
t
e
e
PositivePole
and ve poles
NegativePole
F
84 So polarization is distortion or shifting of e density due to an electricfield nearby In the previous example the neutralatomgot polarized by the cation So cation has a polarizing power and the neutralatom has ability to get polarized ie polarizability Now its the size of atom anion is bigthen its outermost e easy to visualise that
very if cloud is far from the nucleus and loosely bound So it has higher tendency to get polarized ie higher polarizability Whereas in smalleratomsfanions the e cloud is more tightly held
thus making them less polarizable Upon polarization
a dipole is induced in
the moleculeadding to
any permanentdipolealready
present
This polarization caused byexternalinfluence is temporary ie as soon as i remove that external electricfield the atom goes back to original state symmetrical distribution of e cloud
of
85 We define the polarizability of a molecule as the magnitude of the dipoleinduced by one unit of field gradient Bigger the size ofatomlanion smaller the cation Atomic Polarizability
e
higher the polarizing power N S O
C
O 80
loIO
1076
I
higherthe polarizability
7 Br
47
P 3 13
7 U
3.05
s 2.90
2 18
H 0.66
F 7 He 0.55 0.20
asperCRChandbook
A 0.55
polarizability
across a
period
4 down the group
86 Polarizability hasprofound consequences whichyou'll encounter in induced dipole nucleophilicity etc water is a very polarmolecule However methane is much more polarizable than water
This explains the difference b w
µ
fo O
H
I HR
polarity
and polarizability
H
i
II
H
o_0
n
St S E bonds are much more reactivethan C Cl bonds in reactions like
St S
C
t
Lesspolar
Morepolarizable
Alkenes
t
MorePolar
Snf and
EZ
which you will study later
lesspolarizable
arenes with themore E N sp2carbons are lesspolarizablethanalkanes withonly sp3
carbons
Differentmethodsof representing molecularstructures
gy
Wecan represent molecularstructuresthroughdifferent ways lewisstructure structuralformula condensed formula or
line angledrawing
Uts
Ues Ut ut
U
CteCte U Ut Ut3 z
U
Uf
b
te
H
t
Carbon
88
Determine the geometry around
a
Uy
di Ctb
b
all
second row elements in
Ctf Cte OH
c
d
MHz
Predict the indicated bond angles in each compound
a
un H Utz CEC nu
b
Clea
nu
c
Cee
a
e
TI
b
t
I
c
g
µ
N
les C Br
HH
convert each skeletal structure into a complete structure
showing
Utz
89
H
H
les c Br
eachof following compounds
all C's H's and lonepairs d
me
quo N.lk X
h
Determine the geometry around
sp2
a
Ue n
I
116 If Uab
Isps
sp
Trigonal Planar about central carbonatom
by
all
second row elements in
Do
SP
sps
fo N
c
Maginn l te
Tetrahedral around
both C andoxygen
n't
1800
ay
Br
sp2 TrigonalPlanar
H
d
O
H te
H
sp
I
µ
sp
SP3
Tetrahedraland
Tetrahedral
linear
Predict the indicated bond angles in each compound
1800 1200
If b c Cees H as 58421200
A
90
eachof following compounds
H 09.50
I
Uy
fH
Br ale angle are 7109.50 Slight
deviation will be because of different groups
a
ctfu de de
o
Utz
b
l
92
l UI
Utz
I
cee
91
c
cuz
ab
ai
I
ab ii Macey
I
O 11
µ
H
C te
e
ai
okayEmme
Cee
lay
eee
9
h
equine.ie
t
dnue
o
I
H
Ig
Hi
u o
g
cygni H
Ue
o
d
91
OH
te
0
Utz H
orbitals What
a UtzBelt
Predict
the hybridization and
a
I
b
of
geometry aroundeach highlighted atom
d
c
a
92
are used to form each bond in the followingmolecules C Uy O Ctb b W B
Amoxicillin
Nate
Predictthe hybridization and geometryaround each
highlighted atom b Label 5 polar bonds using c
e
Howmany
St and 8
and it bonds does amoxicillinhave
ay
I
µ
11C
1
s overlap
sp
H
c
2
Bee
H
2
sp3 sp overlap Be c
3
s overlap
SP
Be H
H
y
uf
nel care sp Tetrahedral
are
Till LF
ai
O Ctl
00
I
B
ab
sp3
g sp
Ttb
c
sp3
G H
I
H
H H
All care sp2 Trigonal Planar
Ut
Utz
c µ bonds
c te bonds are sp3 s B C bonds are sp2 sp
cee
Tetrahedral
2
e
O
nee cand 0
sp
by
H
b
c Haug 8cg
Cys
sp
c O bonds
H d
HY
SP
sp3
s
sp
Tetrahedral
c
sp linear
µ
sp2
a
f
sp3
sp
P
I
2
sp
Esp2
Polarbonds are highlighted in green bonds 46
IT bonds
6
94
forces 95
Intermolecular
Tries to keep molecules closer between molecules Intermolecular
Intramolecular within same molecule 7 Holds atoms together to make a molecule eg
Intermolecularforces AHect the MetallicBond covalentBond Strength IonicBond
KUMM 400 4000
and chemical properties physical 150 1100 75 1000
molecules point example Nau eg melting boiling Fe Ni Hz Uz Cay
etc reactivity basicity acidity
Cary Ctey etc Ag Pt
behavior the threestate is Physical
Bondingforces ionic covalent metallic are intramolecular because the strength intermere
different are much stronger than intermolecular forces b'coz
cellar forces differfrom stateto state
the e density involved in bonding is veryclosebut in
intermolecular forces thedistance b w the e Ion dipole
density is large
rebond
Dipole dipole
DipoleInduceddipole
Dispersion Londonforces
of
of
of
Intermolecular forces
Ion Dipole forces
96
M be attractive or repulsive but repulsiveforces come into playonly when molecules come so close that their
e clouds start overlapping
forces they are attractiveforces These forcesaffect physical properties St
St
S
S
t St
g
s
8
g g
This is why most time when we discuss intermolecular of
of organic
molecules
St Attractive forces b w cation
polarmolecules
etc
molecules
g
s
St St
St
g
g St
S
S
8
H2O molecule
Nat
8
U
ions are surrounded by H2O St S
mp
eg boilingpoint meltingpoint solubility
simplest example there of forces is when we dissolve
Nacl in water Nat g
bp
U
Attractive forces b w anion
polarmolecules
be
through solubility water
in easily
solids Understood forces
dipole
ion
creditsErebor Mountain
Shutterstock
97 S
ionic of can
St
H
s
u
0
Na
ly si St
S
O
U
st
H
s
O
H
s1
g
O
test
H
H
Watermolecules polar attract the ions from the crystal and hydrate savate
them The forces b w water molecules and ions are won dipole forces
Thetotal energy released during the interaction water and ions is called
of
the HydrationEnergy
Dipoleforces 98
Dipole
These are electrostatic interactions b w moleculeswhich St S St S St S
have permanentdipoles eg Chaz HI MHz HCl etc
oppositely charged ends attractandalign
stronger than London dispersionforces but weakerthan
5 25 KHmol strength ion ion ion dipoleforces as only partial charges are
involved
When oppositely charged ends are aligned the attraction A and P E 14
Nonpolarmolecules don't engage in dipole dipole interactions eg coz Cay etc
strong dipole dipoleforces result in higher boiling meltingpoints and also affect other properties
St St St
S St S S St St S S S H H
H
eg U U U H U H U H G
H H H
I
9
µgc
µgC
µqc
These forces depend on the magnitude
similar Mol wt BP K
ainolar mass
46
Abusing
Uhs o Ue 248
50.48
Uy U 249
Source: chem.libretexts.org
for compounds of
gg
the greater the molecular dipolemoment greater the dipole dipoleforces
Uh 231
ofthe
molecular dipole moment
44 Cee
I
294
41 H
Uy EN 355
100 Hydrogen Bond
Its a special case dipoledipoleforces strongerthan ordinary dipoledipoleforces
It requires H atom bonded to a small highly E N atom specifically 0 N F
S St St S
A H 01h11F B Notany random H can form H bonds Hydrogen which is
Hbond already part a dipole can participate in tebonding with
N O or F St
S S St O H
N H St S S S St S S St S g St S
S g si g g Sf Sf f N H H
St S St
S Ho F H
O
c Hoo o O H X it U also has E.nl comparable to N but
doesn't normally participate in te bonding Nopolarity in this
bond
High E N and small size of NO F create greater St
charge on their covalently bonded H
of
T
of
why
lol
Unique properties Hydrogen that facilitate this uniqueinteraction are electropositivity smallsize
absenceof inner e shells These properties allow X and D to approach eachother closely
without experiencing
X H D any severe repulsion
Intramolecular H bonding tebonding that exists within the same molecule
Intermolecular H H bonding that exists between 2 ditt molecules bonding
ar H D 90
O I O Hoo N µ
e H F I 11 d
o
H H H I o
N XO H O O
µ 0 17 Every water molecule can
para nitrophenol
4 H bonds form
Intermolecular H bonding Intramolecular H bonding
of
g
Neo
If
ie
Enke
no
EIl
102 bonding EHectsofH
Melting and boilingpoints H2O
H2S L Hase
HF 77 HU C HBR PHz
AsHz
Nitze MHz
Urea is
NHS
Hate HI
SbHz
Boiling Point H2O UtzOH 76450 Utz 100C
63C
T
5C
Y
3 H bonds y µ bonds permolecule permolecule
solid while Uts
E Cte
acetone is
Hf CtbUf F ROOM ROH
No H bonding b w ethermolecules
liquid
Amines are more watersoluble than alcohols of comparable mol wt bcoz aminesact as a better donor group in
H bonding solubility
H bonding with water
4 viscosity and
affects acidity
µ
µ
qµ
stronger
H
R
H
Oy
weaker
London Dispersionforces Intermolecular forces 103
So we've discussed the forces between polar molecules but what about the forces that
operate between non polar molecules eg noble gases alkanes Uz Br etc
Intermolecular force blue non polar molecules is Londondispersionforces or weak VanDer
Waal's forces
Theseforces are weakest and exist in all substances polar or nonpolar But since they
are very weak we don't normallymention them in polarsubstances
These forces result from the motion e so they are present between all particles atoms
Ions and molecules Atanyinstant e density on one side atom
of
No inherent polarity can be than the e greater
density on opp in this atom side
But the e in an atom this will create a momentary dipole
are moving randomly all in the atom and this momentary
the time
dipole will polarize the neighbouringatom
far
of
104
This process spreads through all
atoms in the sample
and weak attractive forces b w these
induced dipole come into action These are dispersionforces or weak V D W forces
S
St ooo
Nodipole
ooo
a momentary
ooo
ooo
S
St ooo
9
dipole is created due to randomlymoving e cloud
S
St
Adipole
is induced
in this atom
eventually theseatoms induce dipole in other nearby particles
when particles are close very weak forces and are effective only at low temperature and have less thermal energy to escape from attractiveforces
These are
Londonforces
9 with polarizability
ie biggermolecules or the ones with more surface area have stronger London forces
105 strength
of
London dispersion forces
Polarizability of the particles
I
depends on no
ofe
which is greaterin particle
with highermolar mass Dispersionforces
9 with mol wt
ie
Ho L C Hg L CyHo He L Ne s Ar L Kr L Xe
Chey
Fz
U
Brz
L Iz
Exceptforthe forces b w small highlypolarmolecules or blue molecules forming Hbonds the dispersion force is the dominant intermolecularforce e.g In HCl
85 attraction is dueto dispersionforces and 15
due to dipoledipoleforces
106
what
if two
e.g
molecules non
polar havesame molarmass
like isomers
one with greater surface area has greater dispersion forces n pentane
and
neo pentane
Branchingi reduces surface area
same
molar mass
straightchain has greater surface area among isomers
Neopentane n pentane
Hawk ayub bp 36.1 C
GreaterSurfaceArea
Uts Joab
f
Utz
Utz
bp g 5 C lessersurface area
107
Geckos
How walls on stick Vander Wallsforces
A Gecko's foot has toepadsconsisting of about half a million setae made keratin Eachofthesefinehairs hashundreds of of even smaller projections
protruding
of nanoscale diameterscalledspatulae
fromtheir ends
The VanderWaals forees created b w
the spatulae and the surface In the case of geckofeet the spatulae are
so small and get so close to the surface that an attractive VanderWaals
forceofaround 0 UµN develops b w a singlespatulaand a surface
Credits marimalShutterstock
each substance for identify the
has the higher boilingpoint
MgUz
a
a
Mga
or
Pcl3
consists
of Mg
b
108
keyintermolecularforceIs and predictwhichone ofthe pair CtlNHz Or
Uff
c
UtzOH
b
Utz
and U ions held togetherbyionic bonding forces Pcl
polar molecules so intermolecular dipole dipole forces are present stronger so
or
H
consists
of
Theforces in Mgelz are
it should have a higher boiling point
AGNHz and Uff both consist of polarmolecules ofabout the same molarmass UtsMHz has N H bonds so it can form H bonds but Utz f can'tform H bonds
would
Both
BE or II be expected tohave the higher boilingpoint moleculeshavesame no of e but IU has a dipole moment ofthese
a Can a molecule havedipolemoment
b How is it a No
possible
b
if it has no polarcovalentbonds
for a molecule tohave polar bonds but
If the
various polarbonds are symmetrically placed and their vector sum
down a fewproperties list of water that
High m p
bp
highspecificheat
Which
no dipolemoment
is coming out to be zero then the molecule will be non polar eg coz Cay etc
ay
Og
originate because of
H bonding
highheat of vaporization lowdensityofice compared to water
makeswater a good coolant
of followingmolecules can exhibit
H bonding
etc
a
CtesuezOH
b
Utz o Utz
c ClyNHz
ctbUfOH
it
8
UtzNHz as 2 Hare bonded to N
CMU
and
e
UHH
h
CHU
110
OH
has no H directly bonded to an EN atom but still it shows H bonding Hydrogen atom should have ve polarity on it to participate in H bonding This
Although Clea
bcoz
can also be achieved
eg
U
UTC I
Q
Calculate
f
d LettsgN
OH
a
c atom is bonded to multiple EM atom as in case of Cetus
if
St
H
p
s O
This example shows
1413
c
Hbond
H bonding blue chloroform
ab
Ulcers
and
acetone
the max number of atoms that are planar in the following molecules
Ctf Utz
b
Ctfu
c
Cte
Cte
Utz
d
Benzene
e
Ctf 1
642
a
F pnfjrnee7 mn
1 c
CIC
y H
atoms in
one plane
X Hl
42
C
H
c
by
H
G H
o
Hd
ttI
C BACK
H H
outofPlane H
FRONT
c c
C
c c
c
H
3patfnns.in
H
6 carbons and
6 hydrogens are in sameplane
one
H
e
Ht C
H
All
H
Importance
H
d
y
111
yBackoftheplane
pyke
afonmssanaYepianar
H5
H
pFanne
SH
Z
C u
C s
H
H
b atomsare planar
This moleculebelongsto Allene family
the hydrogens on end carbonatoms in uy why
µ
C
g
TH
C
lµ
The p orbital overlapping
verticalPlane
C CHz
sp has 2purep
orbitalsformaking 2 IT bonds
C
ay d
pz
d d
p2
IT bonds
Py
Utz
O d
Py
In adjacent 1Tbond systems Callenes adjacent ITbonds in 1 planes
C
ie
T
q
for IT bond is 1 to the sp2carbonatoms
4
C
2
Uk are in 1 planes
this C forms 2
1C
Uy
c
942 c Ctez
HorizontalPlane
Uj
c
c
cHz
this figureclearly explains that 21T bonds formed by central Catom have to be in 1 planes c te bonds are present in the plane 1 to the 1T e cloud Utzgroups are in 1 planes
113
Howdoes molecular polarity affect physical properties
of compounds
Molecule having higher polarity
has higher boilingpoint
More energy is required to separatethem
Greater the intermolecular attractive forces
More polarity 9 the solubility in polarsolvents
like dissolves like
less polarity In the solubility in less polar1nonpolar solvents
He
leads to higher melting a boilingpoint
what type
114
of intermolecular forces are present in eachcompound
a
In
b
c
Note
d
e
which compound in each pair has the higher boilingpoint
a
C
e
N
or
or
koµ
7 or
H
b
d
0
Iko
ay
f
dy
or
OH
d
W
or
free
or
M
Az
ace
115
a
c
Koµ
H bonding
d
lesser
dispersionforces
higher bp
e
or
koµ
b
compact
greater Dispersionforces
dipole dipole
Higher bp or
Hbonding GreaterBp
omg
i
Dipole Dipoleforces
Dipole Dipole e
Te
highersurfacearea C
there is some polarity
y µ
or
f
Due to presenceof N
b
Dispersionforces
dipole dipole
a
No polarity
d
f
onlydipoledipole
w
free
h
Dipole Dipoleforces
or
OH
K
q
onlydispersion forces
higher b p
o
or
bonding i Higher b p
Biggermolecule greater dispersionforces
I higher B P
has biggeratom I i
Greater dispersionforces
116
Functional Groups in Organic chemistry
Most c bonds These bonds are strong non polar and not organiccompounds have C and C H
brokeneasily But apartfrom C and H organic compounds sometimes have 0 N S P and
halogens Atoms other than C and H are called Heteroatom
T
different
S OH Cte 692 0 H NH
eog
what differentiates these molecules
to
structural
features like heteroatoms it bonds carbon chain etc
These structural features decide the geometry reactivity physicalproperties molecules
Coote group
eg carboxylicacids are acidic due to
Utz
Utz
Utz
Clb
di
ay
di Utz
I
of
117
Functional the characteristic physical chemical group is a group atoms that is responsible
properties the compound In simple words F G decideshow a particular class compounds
alcohols are due to OH group
react OH is functional group Typical reactions
COOH carboxylicacids are due to COOHgroup
Functional Group
Onecompound can have more than one f G as well R
carbon ReactivePart
N OH µ
Skeleton
OH 0
11
2
Br
HO
CHO
Highlighted groups are AG
of eg
for
of
of
of
T
tycoon
I
f
INCOOR
T
Typeof Compound
GeneralStructure
Example
Alkane
R H
Alkene
4 4
Uez cH Utz
Alkyne
CEC
Hec Cte
AlkylHalide
functional Group
Uy UtzUes c
c c
R X
ay ay
Alcohol
R OH
cte5UyOH
OH CHydroxygroup
Ether
RO R
Ues O Utz
OR Alkoxy group
Amine
R NHz RzNH
UtzMHz
MHz aminogroup
Ryn
G 3N
Thiol
X f U Br I
R SH
118
Br
UtzzMH
Uhs SH
X Halogroup
SH
MercaptoGroup
Sulphide
R S R
Utz Stetz
0 11
O 11 Uys C H
O
O 11 Cee C cute
Aldehyde
r C H
Ketone
11 R C R
Carboxylicacid
11 R C OH
0
0
Esler
11
RC o R
O
11
cage o te O C Cee O Ub 11
O
Amide
R
11 c MHz
O
Acidchloride
11 R c a
O
Utz
11 C
MHz
O
11
chest a
SR alkylthiogroup O
11 c H
die O 11 C
ote
O
11 C o
O 11 C
O
11 C
Y u
carbonylgroup
carboxy group
119
carbon and Hydrogen atoms 120
Types
Based on the number carbonatom connected to a particular carbonatom we classify it as
primary lo secondary120 tertiary 5 or quaternary Cli
which is bonded to only 2 carbonatom
10 or Primary carbon
e.g 20 or secondarycarbon 2 which is bonded to only 2 carbonatoms
C c
I 1 30 or tertiary carbon 2 which is bonded to
c only 3 carbonatoms
10 or Primary U or Quaternary carbon which is bonded to 4 carbonatoms 20orsecondary
lo C C Uy
lo Gt3 l lo Cte f lo 20
I lo I CC CC Ctl c Cc c Uf
l 30 so l
e e
61220 40 orQuaternary
5 or Tertiary
Utz to
of
of
d
d
Ctf
yes quoth Uy ab 1
Degree
CHS
121
of
H atom
Ctf 20
CH 10
Degreeof Catom to which 10 Hydrogen
20 Hydrogen
6
it is
go
2
connected
to
Cle
Utz de 8
µ
Uf Ctl 20
10Hydrogens
g
20 Hydrogens
2
30 Hydrogen
I
122
Solubility
called solvent the extent towhich a compound called the solute dissolves in a
ie Pandit dissolve
Like dissolves like ie polarcompoundsdissolve in polarsolvents and nonpolarsolutes in non polar
solvents
Polar solvent as it has polarity can form H bonds Haler
Organicsolvents Mostly are non polar e.g Cay hexane benzene or are weaklypolar
e
g UtzO Ctb Oil
oil acetone
oil layer
water water
organicsolvent
dissolves oil non polar oil non polar
Nacl ionic
dissolves in doesn't dissolve in water bothare polar organicsolvent nonpolar
in water bcoz
if
liquid
IHT
Nacl
f
f
123 students sometimes wonder that
it
dissolves in water so easily
if
NaCl hasionic bonding which is
and
breaks
strongestthen how come
the ionic bonds b w Nat and I
No doubt that ion ion interactions in ionic solid like NaCl are strong but when added to water each icon is surrounded by so many water molecules through iondipole
interactions Though these forces ion dipole are weaker there are so many ofthem
that they compensate for the stronger ionic bonds H
St
St
H
H OS
St
H
1
l
s 0
8O
1
st
H
St H
O H
1st St l H
H
solvation hydration
o
l
St
ofcation
H
StHl s
s
St
S O
St
S St O µ h
St
H
g1
H og
H
St
l
H ft
Solvation hydration
of anion
Mostorganiccompounds are soluble in organic solvents is water solubleonly
five carbon atoms
if
it contains one polarfunctional group
p
Non polar
Nonpolar Hydrophobic
Anorganiccompound
Nor 0 containing
110
OH
Hydrophilic
waterloving
4
Hydrophobic
waterhating
Hydrophilic
Carbonchain hydrophobic should not be verylong
the substance has to dissolve in water
Cay
Soluble
Insoluble
forevery
polarhead can form H bond with water
if
Cay
f
Polarhead
Tail
Hydrophobic
Butane
124
it contains COOH
water hating
like
dissolveslike
Acetone
f
H2O
Soluble
Soluble
O 11
Acetone has
125
carbons and a polar in functional groups which makes it capable only 3 of H bonding with water In fact acetone is so soluble in water that acetone and water are miscible they form solution in all proportions with each other
s
s1
H
S
o
µ
Ls
0
H
HO
Utz OH
slightlysoluble very in water b'coz carbonchain hydrophobic is small as Cchaingetsbigger
OH
Insoluble in water as c chain is very
big
04
Khao boy
NOH
soluble inwater
µ bond
OH
but
St
note
Glucose
is
far
carbon chain is bigger but we
have 5 OH groups
f Benzoicacid
water Insoluble Bigcarbon chain
we
d
OH
Aceticacid WaterSoluble
That's it
for organic
the
basics of
chemistry
but
dost abhi mere Picture hai See u baaki
in next book